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How do you find #(dy)/(dx)# given #y^2+3x=2#?

Answer 1

Elijah Abram

#(dy)/(dx)=-3/(2sqrt(2-3x))#

As #y^2+3x=2#, we have

#y^2=2-3x# and #y=sqrt(2-3x)#

and using chain rule

#(dy)/(dx)=1/(2sqrt(2-3x))xx(-3)#

= #-3/(2sqrt(2-3x))#

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Answer 2

Emily Ammerman

To find ( \frac{dy}{dx} ) given ( y^2 + 3x = 2 ), differentiate both sides of the equation with respect to ( x ) and solve for ( \frac{dy}{dx} ).

[ \frac{d}{dx}(y^2 + 3x) = \frac{d}{dx}(2) ] [ 2y \frac{dy}{dx} + 3 = 0 ] [ 2y \frac{dy}{dx} = -3 ] [ \frac{dy}{dx} = \frac{-3}{2y} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.